Difference between revisions of "2006 AMC 8 Problems/Problem 18"
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==Problem== | ==Problem== | ||
A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white? | A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white? | ||
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<math> \textbf{(A)}\ \frac{1}{9}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{4}{9}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{19}{27} </math> | <math> \textbf{(A)}\ \frac{1}{9}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{4}{9}\qquad\textbf{(D)}\ \frac{5}{9}\qquad\textbf{(E)}\ \frac{19}{27} </math> | ||
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| − | Solution 2 | + | ==Solution 2== |
| − | We can notice that each face is the same, | + | We can notice that each face is the same, so each face is <math>\boxed{\textbf{(D)}\ \frac{5}{9}}</math> white. |
==See Also== | ==See Also== | ||
Latest revision as of 01:21, 7 January 2020
Contents
Problem
A cube with 3-inch edges is made using 27 cubes with 1-inch edges. Nineteen of the smaller cubes are white and eight are black. If the eight black cubes are placed at the corners of the larger cube, what fraction of the surface area of the larger cube is white?
Solution
The surface area of the cube is 
. Each of the eight black cubes has 3 faces on the outside, making 
black faces. Therefore there are 
white faces. To find the ratio, we evaluate 
.
Solution 2
We can notice that each face is the same, so each face is 
white.
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
